Question: Let $a(x)=x^3+2x^2+x$, and $b(x)=x^2+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{x^3+2x^2+x}{x^2+1}$ : First, we divide ${x^2}$ into ${x^3}$ and get ${x}$ : $ \hphantom{1567|14} {x}\\ {{{x^2}+1}}|\overline{{x^3}+2x^2+x}\\ \hphantom{37.....|}\llap{-}\underline{(x^3+0x^2+x)}\\ \hphantom{37|3.........}+2x^2\\ $ [What did we do here?] Next, we divide ${x^2}$ into ${2x^2}$ to get ${+2}$ : $ \hphantom{1567|14} {x \ {+ \ 2}}\\ {{{x^2}+1}}|\overline{x^3+2x^2+x+0}\\ \hphantom{37.....|}\llap{-}\underline{(x^3+0x^2+x)}\\ \hphantom{37|3.........}{+2x^2}+0x+0\\ \hphantom{37..............|}\llap{-}\underline{(2x^2 + 0x +2)}\\ \hphantom{37|3.............8888........}{-2}\\ $ [What did we do here?] The process stops here because $x^2+1$ is a polynomial of the second degree and $-2$ is a polynomial of the zeroth degree. So it follows that ${r(x)}={-2}$, ${q(x)}={x+2}$, and $ \dfrac{x^3+2x^2+x}{x^2+1}={x+2}+\dfrac{{-2}}{x^2+1}$ To conclude, $q(x)=x+2$ $r(x)=-2$